Problem: $\dfrac{d}{dx}(-x^3+5x^2-8x)=$
According to the sum rule, the derivative of $-x^3+5x^2-8x$ is the sum of the derivatives of $-x^3$, $5x^2$, and $-8x$. The derivatives of these terms can be found using the power rule : $\dfrac{d}{dx}(x^n)=n\cdot x^{n-1}$ For example, this is the derivative of the first term: $\begin{aligned}\dfrac{d}{dx}(-x^3)&=-1\dfrac{d}{dx}(x^3)&&\gray{\text{Constant multiple rule}}\\\\ &=-1 \cdot(3x^2)&&\gray{\text{Power rule}}\\ \\ &=-3x^2\end{aligned}$ Here is the complete differentiation process: $\begin{aligned} &\phantom{=}\dfrac{d}{dx}(-x^3+5x^2-8x) \\\\ &=-1\dfrac{d}{dx}(x^3)+5\dfrac{d}{dx}(x^2)-8\dfrac{d}{dx}(x)&&\gray{\text{Basic differentiation rules}} \\\\ &=-1\cdot3x^2+5\cdot2x-8\cdot 1x^0&&\gray{\text{The power rule}} \\\\ &=-3x^2+10x-8 \end{aligned}$ In conclusion, $\dfrac{d}{dx}(-x^3+5x^2-8x)=-3x^2+10x-8$.